What is the inverse of the function $h(x)=\dfrac{5+x}{4-2x}$ ? $h^{-1}(x) =$
Answer: Let's start by replacing $h(x)$ with $y$. $y=\dfrac{5+x}{4-2x}$ Now let's swap $x$ and $y$ and solve for $y$. $\dfrac{5+y}{4-2y}=x$ [Why do we swap x and y?] $\begin{aligned} \dfrac{5+y}{4-2y}&=x \\\\ 5+y&=x(4-2y) \\\\ 5+y&=4x-2xy \\\\ y+2xy&=4x-5 \\\\ y(1+2x)&=4x-5 \\\\ y&=\dfrac{4x-5}{1+2x} \end{aligned}$ In conclusion, this is the inverse function: $h^{-1}(x)=\dfrac{4x-5}{1+2x}$ [I saw someone solve this problem by originally solving for x. Were they wrong?]